Chapter 12 12.pdf · 12.7 Elasticity: Tension and Compression • For simple tension or...

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Chapter 12 Equilibrium and Elasticity HyperPhysics: Equilibrium Simple Machines

Transcript of Chapter 12 12.pdf · 12.7 Elasticity: Tension and Compression • For simple tension or...

Chapter 12

Equilibrium and Elasticity

HyperPhysics: Equilibrium Simple Machines

12.2 Equilibrium

• What allows objects to be stable in spite

of forces acting on it?

• Under what conditions do objects

deform?

The two requirements for the state of

equilibrium are:

1. The linear momentum of the center of

mass is constant.

2. The angular momentum about the center of

mass, or about any other point, is also

constant.

The balancing rock of Fig. 12-1 is an example of an

object that is in static equilibrium. That is, in this

situation the constants in the above requirements

are zero.

Fig. 12-1 A balancing rock. Although its perch

seems precarious, the rock is in static

equilibrium. (Symon Lobsang/Photis/ Jupiter

Images Corp.)

12.3 The Requirements of Equilibrium

Another requirement for static equilibrium:

12.4 The Center of Gravity

The ladder and the forces on it: Forces in both directions sum to 0:

The torques are all perpendicular

to the plane of the page, so there’s

only one torque equation:

Solve the 3 equations to get:

Force, x: n1 n2 0

Force, y: n1 mg 0

Torque: Ln2 sin L

2mgcos 0

tan sin

cos

1

2

12.4 Example: Static equilibrium, leaning ladder

• At what angle will the ladder slip?

12.4 Example: Static equilibrium

Clicker question The figure shows a person in static equilibrium leaning against

a wall. Which one of the following must be true?

A. There must be a frictional force

at the wall, but not necessarily

at the floor.

B. There must be a frictional force

at the floor, but not necessarily

at the wall.

C. There must be frictional forces

at both wall and floor.

12.4 Example: Static equilibrium

12.4 Example: Static equilibrium, cont.

12.4 Example: Static equilibrium

12.6: Intermediate Structures

12.7: Elasticity

• A stress is defined as deforming force per unit area, which produces a

strain, or unit deformation.

• Stress and strain are proportional to each other. The constant of

proportionality is called a modulus of elasticity.

Stress = Modulus x Strain

12.7 Elasticity: Tension and Compression

• For simple tension or compression, the

stress on the object is defined as F/A,

where F is the magnitude of the force

applied perpendicularly to an area A on

the object.

• The strain, or unit deformation, is then the

dimensionless quantity ΔL/L, the fractional

change in a length of the specimen.

• The modulus for tensile and compressive

stresses is called the Young’s modulus

and is represented in engineering practice

by the symbol E.

Fig. 12-12 A stress–strain curve for a steel test

specimen. The specimen deforms permanently

when the stress is equal to the yield strength of the

specimen’s material. It ruptures when the stress is

equal to the ultimate strength of the material.

12.7: Elasticity: Shearing

• In the case of shearing, the stress is also a force per unit area, but the force

vector lies in the plane of the area rather than perpendicular to it.

• The strain is the dimensionless ratio Δx/L, with the quantities defined as

shown in the figure.

• The corresponding modulus, which is given the symbol G in engineering

practice, is called the shear modulus.

12.7 Elasticity: Hydraulic Stress

• In the figure, the stress is the fluid pressure p on the object, where

pressure is a force per unit area.

• The strain is ΔV/V, where V is the original volume of the specimen

and ΔV is the absolute value of the change in volume.

• The corresponding modulus, with symbol B, is called the bulk

modulus of the material. The object is said to be under hydraulic

compression, and the pressure can be called the hydraulic stress.

12.7: Elasticity

Example: elongated rod

Example: wobbly table

We take the table plus steel cylinder as our system.

The situation is like that in the figure. If the tabletop

remains level, the legs must be compressed in the

following ways: Each of the short legs must be

compressed by the same amount (call it ΔL3) and thus

by the same force of magnitude F3.The single long leg

must be compressed by a larger amount ΔL4 and thus

by a force with a larger magnitude F4.